contestada

Suppose that in a population the frequency of a particular recessive disease is 1/400. assume the presence of only two alleles: dominant allele (a) and a recessive allele (a) in the population and that the population is at hardy-weinberg equilibrium. what is the frequency of the recessive allele that causes the condition?

Respuesta :

mathmate

Given :

Frequency of recessive disease = 1/400

Under Hardy-Weinbert equilibrium, we assume

p = frequency of dominant allele

q= frequency of recessive allele,

 

Then frequency of recessive disease=qq, or

q=sqrt(1/400)=1/20=0.05

=>

p=1-q=1-0.05=0.095

 

Answer: frequency of the recessive allele, q, is 1/20, or 0.05, or 5%

Using Hardy-Weinberg equilibrium, we have

p = frequency of dominant allele

q= frequency of recessive allele,

Then frequency of recessive disease=qq, or

q = sqrt(1/400) = 1/20= 0.05

p = 1-q = 1-0.05 = 0.095

Frequency of the recessive allele, q, is 1/20, or 0.05, or 5%

What is Hardy-Weinberg equilibrium ?

Hardy–Weinberg Equilibrium (HWE) is a null model of the relationship between allele and genotype frequencies, both within and between generations, under assumptions of no mutation, no migration, no selection, random mating, and infinite population size.

What are the 5 principles of the Hardy-Weinberg equilibrium?

There are five basic Hardy-Weinberg assumptions: no mutation, random mating, no gene flow, infinite population size, and no selection.

To learn more about Hardy–Weinberg Equilibrium, here

https://brainly.com/question/419732?referrer=searchResults

#SPJ2

Otras preguntas